A) 6.2
B) 4.12
C) 3.09
D) 7
Correct Answer: A
Solution :
\[6.02\times {{10}^{22}}\] molecules of each \[{{N}_{2}},{{O}_{2}}\] and \[{{H}_{2}}\] \[=\frac{6.02\times {{10}^{22}}}{6.02\times {{10}^{23}}}\] moles of each Weight of mixture = weight of 0.1 mole \[{{N}_{2}}+\] weight of 0.1 mole \[{{H}_{2}}\]+ weight of 0.1 mole of \[{{O}_{2}}\] \[=(28\times 0.1)+(2\times 0.1)+(32\times 0.1)\] \[=6.2gm\]
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